Question

Given: x2 - y2 - 4x - 6y - 6 = 0. Write the standard form of the hyperbola.

1) Find the center

2) Find the vertices

3) Find the foci

4) Find the slope of the asymptotes

5) Determine whether the transverse access is vertical or horizontal.

Answer 1

\[(x ^{2}-4x)-(y ^{2}+6y)=6\]
\[(x ^{2}-4x+4-4)-(y ^{2}+6y+9-9)=6\]
\[(x-2)^{2}-(y+3)^{2}-4+9=6\]
\[(x-2)^{2}-(y+3)^{2}=1\]

Answer 2

great so the center is (2, -3)?

Answer 3

yes

Answer 4

could you please explain how to find the other stuff? im lost

Answer 5

ok

Answer 6

??

Answer 7

you can convert it to standard form by putting::
x=X+2 and y=Y-3

you get:
\[X ^{2}-Y ^{2}=1\]

Answer 8

how do you find the vertices?

Answer 9

the coordinates of vertices wrt to new axes are (\[(X=\pm a,0)ie(X=\pm 1,0)\]

Answer 10

sorry

Answer 11

its ok i understand

Answer 12

\[(X=\pm a,Y=\pm 0)\]
\[x-2=\pm 1,y+3=0\]
so the vertices are (3,-3) and (1,-3)

Answer 13

foci are \[(X=\pm ae,Y=0) \]
where X=x-2 and Y=y+3

Answer 14

e stands for eccentricity

Answer 15

\[e=\sqrt{1+(b ^{2}/a ^{2})}\]

Answer 16

got till here????

Answer 17

yes

Answer 18

asymptotes are::
\[Y=\pm (b/a)X\] where X=x-2 and Y=y+3

Answer 19

ok

Answer 20

slope:
-(coefficient of x)/(coefficient of y)

Answer 21

thanks!

Answer 22

equation of transverse axis is y=0

Answer 23

@kenneyfamily